"solve the following equation for x"

4^(3x-2) = 26^(x+1)

i use logs and get to :

(3x-2)log4 = (x+1)log26 ...but not sure where to go from here...simple I imagine.

thanks

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- Jul 21st 2010, 05:38 AMporge111simple log question
"solve the following equation for x"

4^(3x-2) = 26^(x+1)

i use logs and get to :

(3x-2)log4 = (x+1)log26 ...but not sure where to go from here...simple I imagine.

thanks - Jul 21st 2010, 05:49 AMHallsofIvy
"log 4" and "log 26" are just numbers. If it were (3x- 2)A= (x+ 1)B for A and B any numbers, you would multiply to get 3Ax- 2A= Bx+ B. Subtract Bx from both sides and add 2A to both sides to get (3A- B)x= B+ 2A. Finally, (as long as ) divide both sides by 3A- B to get . Now, replace A by log 4 and replace B by log 26.

- Jul 21st 2010, 05:58 AMProve It

. - Jul 21st 2010, 11:41 AMSoroban
Hello, porge111!

Quote:

Solve for

i used logs and get to: .

... but not sure where to go from here.

You have: .

. . . . . . .

Factor: .

Therefore:. . . . . . . . . . . .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This can be simplified beyond all recognition . . .

. .

. .

. .

- Jul 21st 2010, 11:05 PMProve It
- Jul 22nd 2010, 08:45 AMporge111
thanks guys!

I was just being silly as usual it seems.